Find all points of discontinuity of f , where f is defined by f ( x ) = { x 2 + 3 , if x 3 x 2 - 3 , if x 3

Question

Find all points of discontinuity of $f$ , where $f$ is defined by

$f (x) =$ ${\begin{matrix} x^{2} + 3, & if x \leq 3 \\ x^{2} - 3, & if x > 3 \end{matrix}$

Smart Achievers · Accepted Answer

(2)

Ans. $f$ is discontinuous at $x = 3$

Explanation:

The given function $f$ is

$f (x) =$ ${\begin{matrix} x^{2} + 3, & if x \leq 3 \\ x^{2} - 3, & if x > 3 \end{matrix}$

$(i) c < 3 (ii) c = 3 (iii) c > 3$

where $c$ be a point on the real line. Then these three cases are valid.

Case (i): $c < 3$

Then, $f (x) = x^{2} + 3$

$f (c) = c^{2} + 3$

$\lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} + 3) = c^{2} + 3$

$∴$ $\lim_{x \to c} f (x) = f (c)$

Therefore, $f$ is continuous at all points $x$ , such that $x < 3$ .

Case (ii): $c = 3$

Then, $f (x) = x^{2} + 3$

$\lim_{x \to c^{-}} f (x) = \lim_{x \to 3^{-}} (x^{2} + 3) = 3^{2} + 3 = 12$

The value of the function at $c = 3$ is $f (3) = 3^{2} + 3 = 12 .$

$\lim_{x \to c^{+}} f (x) = \lim_{x \to 3^{+}} (x^{2} - 3) = 3^{2} - 3 = 6$

Since $\lim_{x \to 3^{-}} f (x) \neq \lim_{x \to 3^{+}} f (x),$ the limit does not exist.

Hence, $f$ is discontinuous at $c = 3$ .

Case (iii): $c > 3$

Then, $f (x) = x^{2} - 3$

$f (c) = c^{2} - 3$

$\lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} - 3) = c^{2} - 3$

$∴$ $\lim_{x \to c} f (x) = f (c)$

Therefore, $f$ is continuous at all points $x$ , such that $x > 3$ .

$∴$ The given function $f$ is discontinuous at point $c = 3$ .

Want Us to Email you About Special Offers & Updates?